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var a=[1,2,3,4,5,6];
6 B2 h8 b. m: Rvar b=[2,3,6];
% L. [0 J. G, ]/ p9 F8 j! ]7 qvar c =[];1 U' P1 r$ z: D& c2 b
for(var i=0;i<a.length;i++){
# \- p7 i/ q# D- \0 avar has = false;$ b6 v8 Z8 r; z, [& I2 x8 G
var data = a[i];5 I" J% I& J& ]% f
//判断baic中是否有du元素data. @. A* y% a1 b {' m
for(var j=0;j<c.length;j++){
: t5 v, r* D5 x+ \( gif(data ==c[j]){; N, h6 j4 U: o# z, N M- _0 I
has =true;/ ~+ m* h( A! s; U% ?; c* n* u' H
break;
2 {. Q$ a" s7 k2 x2 v}
- X, B/ l8 H) j4 g}
4 s- Q7 a1 _7 T4 r6 n: u//如果zhi没有. S; f; ~3 }' w# J C S) B
if(has==false){
, X/ p! Q$ x. \( J8 N$ Xc.push(data);2 J! r5 U* ~1 @# R6 n2 R. ~4 ~- a! V
}2 i4 @: o9 W9 K" i
}
! L$ u K" U) w/ rfor(var i=0;i<b.length;i++){
: P2 n8 _0 l3 e) A* ~. X9 lbool has = false;4 P9 o% B& K6 S2 f* U E
var data = a[i];7 Q2 N4 ~' ~) m. @; C: Z
//判断c中是否有元素data
p& @0 o. B& L, J" R( }for(var j=0;j<b.length;j++){
; F+ ]+ K/ b! k, d* l0 Dif(data ==b[j]){. r) h2 X) t% A. g T+ t% i$ _: j
has =true;3 {# ^1 |+ ?; t7 Q" L
break;
& {( [3 i1 e' c* f/ ]}. u. |% [% W% I* F8 W( W
}5 m' o: H% U/ t* @5 ~. m
//如果没有5 ]2 e: P4 P5 q! l
if(has==false){
. b! l1 ?: } i/ d1 O8 Yc.push(data);
" I1 Z/ P4 o# L8 K- z}
' O' k* ?$ c' |+ Y1 G}7 W! u; W5 f5 W" D2 i7 u" x
//最后c就是结果 |
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